College Algebra

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Open Chapter Ch. 1: Graphs, Functions, and Models
Lesson #1 Graphing
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Lesson #2 Graphing Functions
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Lesson #3 Slope of Linear Functions
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Lesson #4 Linear Equations and Modeling
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Lesson #5 Linear Equations, Functions, Zeros and Applications
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Lesson #6 Linear Inequalities
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Open Chapter Ch. 2: More on Functions
Lesson #7 Analyzing Functions
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Exam Exam 1
Lesson #8 Algebra for Functions
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Lesson #9 The Composition of Functions
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Lesson #10 Symmetry
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Lesson #11 Transformations
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Lesson #12 Equations of Variation
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Open Chapter Ch. 3: Quadratic Functions and Equations; Inequalities
Lesson #13 Complex Numbers
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Lesson #14 Quadratic Equations, Functions, Zeros and Models
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Lesson #15 Analyzing Graphs of Quadratic Functions
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Lesson #16 Rational and Radical Equations
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Exam Midterm Exam
Lesson #17 Absolute Value Equations and Inequalities
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Open Chapter Ch. 4: Polynomial Functions and Rational Functions
Lesson #18 Polynomial Functions and Models
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Lesson #19 Graphing Polynomial Functions
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Lesson #20 Polynomial Division
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Lesson #21 The Zeros of Polynomial Functions
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Lesson #22 Rational Functions
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Lesson #23 Polynomial and Rational Inequalities
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Open Chapter Ch. 5: Exponential Functions and Logarithmic Functions
Lesson #24 Inverse Functions
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Lesson #25 Exponential Functions and Graphs
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Lesson #26 Logarithmic Functions and Graphs
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Exam Exam 3
Lesson #27 Properties of Logarithmic Functions
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Lesson #28 Solving Exponential Equations and Logarithmic Equations
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Lesson #29 Applications and Models: Growth and Decay; Compound Interest
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Open Chapter Ch. 6: Systems of Equations
Lesson #30 Systems of Equations in Two Variables
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Lesson #31 Systems of Inequalities and Linear Programming
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Lesson #32 Nonlinear Systems of Equations and Inequalities
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Lesson #33 Sequences and Series
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Exam Final Exam

Assignments:

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Lesson Objectives:

- Graph linear inequalities.
- Graph systems of linear inequalities.
- Solve linear programming problems.


The graph of an inequality represents all of its solutions.

Graph `y >= 7`.

So start with the related equation, `y = 7`, which is a horizontal line which goes through the point (0, 7). And then, since `y >= 7`, we're going to shade above the line. The solution to the inequality, `y >= 7`, includes all of the points on or above the line, `y = 7`.



Systems of linear inequalities consist of two or more linear inequalities considered simultaneously.

To produce a graph, solve for y and plot the related equation of the line.

Then determine where to shade for each inequality, by choosing a test point. Plug the point into the inequality, and if it yields a true statement, then shade in the region containing the test point.

The region where all of the shading overlaps from all of the inequalities is the solution.

The points at which the lines intersect are called vertices.



So here we're given a system of three inequalities.

We'll start with our first inequality, `x-2y <= 2`. So start by solving for y. We'll subtract x from both sides, and we get `-2y <= 2-x`, and then divide both sides by -2. Don't forget to flip the sign. So `y >= -1+1/2x`.



Now we need to go ahead and graph the related equation. That is, `y = 1/2x-1`. So the y-intercept occurs at the point (0, -1), and then the slope is `1/2`. Rise 1, over 2. So at the point (2, 0), we have the x-intercept. And it's a solid line, because the inequality includes the equal sign.

Now if we allow (0, 0) to be our test value, we need to plug into our inequality and see if we get a true statement.

So `0 >= -1+1/2*0`, or `0 >= -1`. Since this is a true statement, we know that we are going to shade in the region which contains the point (0, 0).

Now for our second inequality, `x+y >= 8`.

If we solve for y, we get `y >= 8-x`. The related equation is `y = -x+8`. So our y-intercept occurs at the point (0, 8), and we have a slope of -1. So we have another point at (1, 7). And we have another solid line, because our inequality includes the equal sign.

Now let's go ahead and let (0, 0) be our test value again. So we'll need to plug that into our inequality.



So we have `0 >= 8-0`, or `0 >= 8`. This is a false statement, so that means we are going to shade in the region that does not contain this point. So we're going to shade above the line.

Now for our third inequality, `y-5 <= 0`. So if we add 5 to both sides, we have `y <= 5`. So if we graph the related equation, `y = 5`, our y-intercept occurs at (0, 5). So we're going to draw a horizontal line. Now we can choose our test value as 0 again, and plug it into our inequality. We have `0 <= 5`. Since this is a true statement, we're going to shade in the region which contains our test value.

The solution to our system of inequalities is where all of our shaded regions overlap. That occurs in this triangular region right here. And our vertices are where our related equations intersect.



So let's go ahead and solve for our first vertex, where `y = 5` and `y = 1/2x-1` intersect, by setting our equations equal to each other. So `5 = 1/2x-1`. Add 1 to both sides, `6 = 1/2x`. Multiply both sides by 2, and we have `12 = x`. So our first vertex occurs at the point (12, 5).

And then for our second vertex, where `y = 5` and `y = -x+8` intersect, we need to set those equations equal to each other. `5 = -x+8`. Subtract 8 from both sides, and we have `-3 = -x`, or `x = 3`. So our second vertex occurs at the point (3, 5).

Now for our third vertex, we need to set `y = 1/2x-1` equal to `y = -x+8`. `1/2x-1 = -x+8`. Add 1 to both sides, we have `1/2x = -x+9`. And then add x to both sides, and we get `3/2x = 9`. And then if we multiply both sides by 2/3, 9 goes into 3, 3 times, and `3*2 = 6`, so `x = 6`. And then plugging 6 into either one of our equations will give us `-6+8`, which is 2. So our third vertex is at (6, 2).

So our solution is the area where all the shading overlaps, and our vertices are (3,5), (12, 5), and (6, 2).



In business, linear program is used to maximize profit and minimize cost. Here's an example.

Katie owns a midsize sedan and a smart car. She can afford 14 gallons of gasoline to be split between the sedan and smart car. Katie's midsize sedan gets 30 miles per gallon, and with the fuel currently in her tank, can take at most 10 more gallons of gas. Her smart car gets 80 miles per gallon and can hold at most 7 more gallons of gas. How many gallons of gas should each vehicle use if Katie wants to travel as far as possible on the 14 gallons of gas? And what is the maximum number of miles that she can travel?

So let's allow x to represent the gas in our sedan, and y represent the gas in our smart car. Since we're going to buy at most 14 gallons of gas for both of our vehicles, `x+y <= 14`.

So start by graphing the related equation, `y = 14-x`. So we have the y-intercept at (0, 14); down 1, over 1 for the slope, and it's a solid line, because the inequality includes the equal sign.

Now if we pick a test value, (0, 0), `0+0 = 0`, and it is less than or equal to 14, so we include that test value in our shading. Now since it says that our sedan can take at most 10 more gallons of gas, the amount of gas that we add to our sedan will be between 0 and 10. So we'll draw a vertical at `x = 0`, and a vertical line at `x = 10`. So x is between 0 and 10. So we'll shade between `x = 0` and `x = 10`.

So then it says that her smart car can hold at most 7 gallons of gas, so y is between 7 and 0. So we'll draw a horizontal line at `y = 0`, and a horizontal line at `y = 7`. And then we shade between the two lines.

So all of our shading overlaps in this region. So we have 5 vertices. One at the point (0, 7), another at the point (0, 0), then (10, 0), (10, 4), and (7, 7). And now, in order to determine the maximum number of miles she can travel, we'll need to put together our miles traveled equation and plug each of our vertices in to see which one has the largest value.

So since she has 30 miles per gallon for her sedan, and 80 miles per gallon for her smart car, `30x+80y` is going to equal the total miles traveled. So now we need to find out what M is for each of our vertices. At (0, 7), we have `30(0)+80(7)`, which is a total of 560 miles. And then at (7, 7), we have `30(7)+80(7)=770" miles"`. And then for (10, 4), `30(10)+80(4) = 620" miles"`. And then for (10, 0), `30(10)+80(0)` gives us 300 miles, and then for (0, 0), we get 0 miles.

So the maximum number of miles she can travel is 770 miles, if she add 7 gallons of gas to her sedan, and 7 gallons of gas to her smart car.